package com.lgc.leetcode.easy.p21mergeTwoLists;

public class Program {
    //    将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。
//    示例：
//    输入：1->2->4, 1->3->4
//    输出：1->1->2->3->4->4
    public static void main(String[] args) {
        //[2,3]
        ListNode l1 = new ListNode(2);
        ListNode curr2 = new ListNode(5);
        l1.next = curr2;
//        ListNode curr3 = new ListNode(5);
//        curr2.next=curr3;

        //[4,5,6]
        ListNode l2 = new ListNode(4);
        ListNode currNew2 = new ListNode(5);
        l2.next = currNew2;
        ListNode currNew3 = new ListNode(6);
        currNew2.next = currNew3;

        Program program = new Program();
//        ListNode resultListNode1 = program.mergeTwoLists(l1, l2);

        ListNode resultListNode1 = program.merge_recursion(l1, l2);

        while (resultListNode1 != null) {
            System.out.println(resultListNode1.val + " -> ");
            resultListNode1 = resultListNode1.next;
        }

        System.out.println("------------------------");


        int a = 0;
    }

    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) {
            return l2;
        }

        if (l2 == null) {
            return l1;
        }

        ListNode head = new ListNode(0);
        ListNode cur = head;

        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                cur.next = l1;
                cur = cur.next;
                l1 = l1.next;
            } else {
                cur.next = l2;
                cur = cur.next;
                l2 = l2.next;
            }
        }

        // 任一为空，直接连接另一条链表
        if (l1 == null) {
            cur.next = l2;
        } else {
            cur.next = l1;
        }

        return head.next;
    }

    /**
     * 递归实现
     *
     * @param l1
     * @param l2
     * @return
     */
    public ListNode merge_recursion(ListNode l1, ListNode l2) {
        if (l1 == null) {
            return l2;
        }

        if (l2 == null) {
            return l1;
        }

        if (l1.val < l2.val) {
            l1.next = merge_recursion(l1.next, l2);
            return l1;
        } else {
            l2.next = merge_recursion(l1, l2.next);
            return l2;
        }
    }
}
